What is the solution for t*y^'=y+((1/4)*t^2+y^2)^{(1/2)} ?

The solution for t*y^'=y+((1/4)*t^2+y^2)^{(1/2)} is y=-1/(4c_{1)}+(c_{1}t^2)/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

ty^'=y+((1/4)t^2+y^2)^{(1/2)}

t \cdot y^{^{'}} = y + ((\frac{1}{4}) \cdot t^{2} + y^{2})^{(\frac{1}{2})}

y = - \frac{1}{4 c _{1}} + \frac{c _{1} t ^{2}}{4}

Solution steps

t \cdot y^{'} (t) = y + ((\frac{1}{4}) \cdot t^{2} + y^{2})^{(\frac{1}{2})}

Solve using substitution:

y = - \frac{1}{4 c _{1}} + \frac{c _{1} t ^{2}}{4}

y = - \frac{1}{4 c _{1}} + \frac{c _{1} t ^{2}}{4}

\int \frac{1}{x 9 - 16 x ^{2}} d x

\int (x^{2} + 2 x + 4) d x

\int_{\frac{x}{z}}^{1} 1 d y

- 5 e^{x + 3} + e^{1}

\frac{\partial}{\partial x} (\frac{1}{2 π} (\frac{y}{x})^{\frac{1}{2}})

What is the solution for t*y^'=y+((1/4)*t^2+y^2)^{(1/2)} ?
The solution for t*y^'=y+((1/4)*t^2+y^2)^{(1/2)} is y=-1/(4c_{1)}+(c_{1}t^2)/4